Delta G standard for keq=1

[umdnjmed]

So....I've gone over this time without number and I think I've finally gotten it so please confirm or correct my understanding.

delta G is always zero at equilibrium, even for standard state conditions.

delta G standard never determines the equilibrium conditions for non standard state conditions. They will always have their own delta G which will determine Keq.

The only concept I have trouble grasping is delta G standard being equal to zero and yielding a Keq of 1. If delta G standard is equal to zero (for a reaction in standard state) how does the reaction get to equilibrium with no driving force? unless you start out with 1M reactant and 1M product

Also, I recall one of my professors saying one of the delta G's was zero for diffusion but i can't remember if it was delta G standard or regular delta G. I think the answer to this second question may be the key to answering the first.

Any help would be greatly appreciated.

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[umdnjmed]

One more question, can an energetically unfavorable reaction be made favorable by increasing reactant concentration?

If for instance you have an exergonic reaction; in the energy diagram, the product ends up with lower energy and this is favorable. In the reverse reaction, the product (which was originally the reactant) is at higher energy, yet the reverse reaction occurs in order to establish equilibrium. Wouldn't this mean at a high enough concentration, an unfavorable reaction (the reverse reaction in this case) can be favorable (and in this case maybe even spontaneous, since external activation energy is not added)?

 

[chiddler]

Not sure if I understand your first question. If Keq is 0, then this just means equilibrium drives the reaction until concentration of reactants = concentration products. if G = 0 means that the reaction is at equilibrium and there will be no conversion

ΔG = ΔG°' - RTln(Q)
and
ΔG°' = -RTln(Keq)

ΔG of reaction depends on the conditions that the reaction is in. If you change the concentrations, then you change ΔG.

One more question, can an energetically unfavorable reaction be made favorable by increasing reactant concentration?

If for instance you have an exergonic reaction; in the energy diagram, the product ends up with lower energy and this is favorable. In the reverse reaction, the product (which was originally the reactant) is at higher energy, yet the reverse reaction occurs in order to establish equilibrium. Wouldn't this mean at a high enough concentration, an unfavorable reaction (the reverse reaction in this case) can be favorable (and in this case maybe even spontaneous, since external activation energy is not added)?

yes, by the above equations.

 

[chiddler]

hey i forgot: what does the apostrophe indicate in ΔG°' as opposed to ΔG°?